Question

A polytropic process for an ideal gas is represented by equation $P{V}^{\alpha }=\text{constant}$. If $\gamma$ is the ratio of specific heats $\frac{C_p}{C_v}$ , then the value of α for which molar heat capacity of the process is negative, is given by

• A. $\gamma >\alpha$
• B. $\gamma >\alpha >1$
• C. $\alpha >\gamma$
• D. none of these

Answer 1

The correct option is B $\gamma >\alpha >1$

In a polytropic process the molar heat capacity of the process doesnot vary.
Given $P{V}^{\alpha }=constant$
We need to find of values of $\alpha$ for which the molar heat capacity is negative.
By the definition of molar specific heat capacity, we get
$\Delta Q=nC\Delta T$
$C=\frac{1}{n}\frac{\Delta Q}{\Delta T}=\frac{1}{n}\frac{dQ}{dT}$ (For instantaneous value)
Using first law of thermodynamics, $dQ=dU+dW$ in the above equation,
$C=\frac{1}{n}\frac{dU+dW}{dT}$
Substituting, $dU=n{C}_{v}dT$ and $dW=PdV$
$C=\frac{1}{n}(\frac{n{C}_{v}dT}{dT}+\frac{PdV}{dT})$
$=C_v+\frac{PdV}{ndT}$....(i)

Using the ideal gas equation, $PV=nRT$
Differentiaing, $\frac{PdV+VdP}{R}=ndT$
Substituting this in equation (i) we have,
$C=C_v+\frac{PdV}{PdV+VdP}R$
$C=C_v+\frac{R}{1+\frac{VdP}{PdV}}$....(ii)

From the given equation,$P{V}^{\alpha }=k\Rightarrow P=\frac{k}{v^{\alpha }}$
Differenciating, $\frac{dP}{dV}=\frac{-\alpha k}{V^{\alpha +1}}$
Substituting this in equation (ii) we have,
$C=C_v+\frac{R}{1+\frac{VdP}{PdV}}=C_v+\frac{1}{1+\frac{V}{k}\frac{(-k)\alpha }{V^{\alpha +1}}{V}^{\alpha }}$
In general $C_v=\frac{R}{\gamma -1}$
$C=\frac{R}{\gamma -1}+\frac{R}{1-\alpha }$
We want this value to be negative,
$\therefore C=\frac{R}{\gamma -1}+\frac{R}{1-\alpha }<0$
Simplyfing the above equation, $\frac{(\alpha -\gamma )}{(\gamma -1)(\alpha -1)}<0$
$\gamma -1$ is positive.
$\therefore \frac{(\alpha -\gamma )}{(\alpha -1)}<0$
Case 1: $(\alpha -\gamma )<0$ and $(\alpha -1)>0$
$\alpha <\gamma$ and $\alpha >1\Rightarrow \gamma >\alpha >1$

Case 2:$(\alpha -\gamma )>0$ and $(\alpha -1)<0$
$\alpha >\gamma$ and $\alpha <1$
This possibility is not possible as $\gamma$ cannot be less than $1$

$\therefore \gamma >\alpha >1$ is correct.